Question

the sum of the digits of two digit number is 8 and the difference between the number and that formed by reversing the digit is 18 find the number

Answer 1

Let the unit place digit be x

And tens place digit be y

No. Formed by x and y = 10y+x (I have multiplied y by ten because it is on tens place)

Now, no. Obtained by reversing digits = 10x+y

Here sum of digits = 8 (I)

And difference on reversing digits is 18

A/q

(10y+x) - (10x+y) = 18

Or, 10y+x-10x-y= 18

Or, 9y-9x = 18

Or, 9(y-x) = 18

Or y-x = 18/9= 2(ii)

On adding both the given equation

Now, y+x = 8

y-x = 2

------------------

2y. = 10

Or, y =5

Now putting the value of y in equation 1st

y+x=8

5+x=8

Or, x= 3

Now the formed no. = 53

Answer 2

Let the digit at unit's place be x and the digit at ten's place be y.

Therefore,

Number = 10y + x

Number formed by reversing the digits = 10y + x

★ According to givEn condition,

The sum of the digits of two digit number is 8.

➟ x + y = 8 [eq. (1)]

And,

The difference between the number and that formed by reversing the digit is 18.

➟ (10y + x) - (10x + y) = 18

➟ 10y + x - 10x - y = 18

➟ 9y - 9x = 18

➟ 9(y - x) = 18

➟ y - x = 18/2

➟ y - x = 2 [eq. (2)]

✇ Subtracting eq. (2) from eq. (1), we get,

⠀⠀⠀⠀⠀⠀⠀ x + y = 8

⠀⠀⠀⠀⠀⠀⠀- x + y = 2

⠀⠀⠀⠀⠀⠀⠀ + ⠀- -

⠀⠀⠀⠀⠀ ___________

⠀⠀⠀⠀⠀⠀⠀ 2x = 6

⠀⠀⠀⠀ ⠀___________

➟2x = 6

➟ x = 6/2

➟ x = 3

✇ Now, Putting value of x in eq. (1),

➟ 3 + y = 8

➟ y = 8 - 3

➟ y = 5

Therefore,

The required number is -

➟ 10y + x

➟ 10 × 5 + 3

➟ 50 + 3

➟ 53

∴ Hence, The required number is 53.